Researchers studying catfish estimated the number of fingerling catfish and large catfish living in different rivers throughout the country. The following histograms summarize the relative frequency for each type of catfish.
The figure presents two histograms. The first histogram is titled Distribution of Fingerling Catfish and has a horizontal axis labeled Number per River, and the numbers 2,000 through 12,000, in increments of 2,000, are indicated. The vertical axis is labeled Relative Frequency, and the numbers 0 through 0.35, in increments of 0.05, are indicated. The data represented in the histogram are as follows. Note that all values are approximate. 2,000 through 3,000 catfish, 0. 3,000 through 4,000 catfish, 0.07. 4,000 through 5,000 catfish, 0.15. 5,000 through 6,000 catfish, 0.04. 6,000 through 7,000 catfish, 0.10. 7,000 through 8,000 catfish, 0.27. 8,000 through 9,000 catfish, 0.07. 9,000 through 10,000 catfish, 0.04. 10,000 through 11,000 catfish, 0.26. 11,000 through 12,000 catfish, 0. The second histogram is titled Distribution of Large Catfish. The horizontal axis is labeled Number per River, and the numbers 250 through 2,250, in increments of 250, are indicated. The vertical axis is labeled Relative Frequency, and the numbers 0 through 0.35, in increments of 0.05, are indicated. The data represented in the histogram are as follows. Note that all values are approximate. 250 through 500 catfish, 0.20. 500 through 750 catfish, 0.18. 750 through 1,000 catfish, 0.18. 1,000 through 1,250 catfish, 0.15. 1,250 through 1,500 catfish, 0.15. 1,500 through 1,750 catfish, 0.07. 1,750 through 2,000 catfish, 0.03. 2,000 through 2,250 catfish, 0.03.
Based on the histograms, which of the following is the best comparison of the means and the ranges for the two distributions?
The mean and range of the fingerling catfish are both equal to those of the large catfish.
A
The mean and range of the fingerling catfish are both less than those of the large catfish.
B
The mean and range of the fingerling catfish are both greater than those of the large catfish.
C
The mean of the fingerling catfish is equal to that of the large catfish, and the range of the fingerling catfish is greater than that of the large catfish.
D
The mean of the fingerling catfish is equal to that of the large catfish, and the range of the fingerling catfish is less than that of the large catfish.
E
While the range is for large catfish is 2000 and the fingerling range is 8000, but how do you find the mean on the histogram?
To find the mean of a distribution represented by a histogram, you need to calculate the midpoint of each class interval and then multiply it by its corresponding relative frequency. The sum of these products will give you the mean.
For the Distribution of Fingerling Catfish histogram:
Midpoint of class interval 2,000-3,000 = (2,000 + 3,000) / 2 = 2,500
Midpoint of class interval 3,000-4,000 = (3,000 + 4,000) / 2 = 3,500
And so on...
Next, you multiply each midpoint by its corresponding relative frequency and sum up the products:
Mean = (2,500 * 0) + (3,500 * 0.07) + (4,500 * 0.15) + (5,500 * 0.04) + (6,500 * 0.10) + (7,500 * 0.27) + (8,500 * 0.07) + (9,500 * 0.04) + (10,500 * 0.26) + (11,500 * 0)
= 0 + 245 + 675 + 220 + 650 + 2025 + 595 + 380 + 2730 + 0
= 8,920
For the Distribution of Large Catfish histogram, you perform the same calculations:
Mean = (375 * 0.20) + (625 * 0.18) + (875 * 0.18) + (1,125 * 0.15) + (1,375 * 0.15) + (1,625 * 0.07) + (1,875 * 0.03) + (2,125 * 0.03)
= 75 + 112.5 + 157.5 + 168.75 + 206.25 + 113.75 + 56.25 + 63.75
= 953.75
Based on these calculations, the mean of the fingerling catfish distribution is 8,920 and the mean of the large catfish distribution is 953.75. Therefore, we can conclude that the mean of the fingerling catfish is greater than that of the large catfish.
Therefore, the correct answer is:
The mean of the fingerling catfish is equal to that of the large catfish, and the range of the fingerling catfish is greater than that of the large catfish. (Option D)
To find the mean on a histogram, you need to identify the midpoint of each bar and calculate the average of these midpoints.
For the Distribution of Fingerling Catfish histogram, you can find the midpoint of each bar by taking the average of the upper and lower limits of each bar.
Here are the midpoints for each bar:
- 2,000-3,000 catfish: midpoint = (2,000 + 3,000)/2 = 2,500
- 3,000-4,000 catfish: midpoint = (3,000 + 4,000)/2 = 3,500
- 4,000-5,000 catfish: midpoint = (4,000 + 5,000)/2 = 4,500
- 5,000-6,000 catfish: midpoint = (5,000 + 6,000)/2 = 5,500
- 6,000-7,000 catfish: midpoint = (6,000 + 7,000)/2 = 6,500
- 7,000-8,000 catfish: midpoint = (7,000 + 8,000)/2 = 7,500
- 8,000-9,000 catfish: midpoint = (8,000 + 9,000)/2 = 8,500
- 9,000-10,000 catfish: midpoint = (9,000 + 10,000)/2 = 9,500
- 10,000-11,000 catfish: midpoint = (10,000 + 11,000)/2 = 10,500
- 11,000-12,000 catfish: midpoint = (11,000 + 12,000)/2 = 11,500
Now, calculate the mean by finding the weighted average of the midpoints using their respective relative frequencies.
(mean) = (midpoint1 x frequency1 + midpoint2 x frequency2 + ... + midpointn x frequencyn)/(sum of all frequencies)
For example, for the Distribution of Fingerling Catfish histogram:
(mean) = (2500 x 0 + 3500 x 0.07 + 4500 x 0.15 + 5500 x 0.04 + 6500 x 0.10 + 7500 x 0.27 + 8500 x 0.07 + 9500 x 0.04 + 10500 x 0.26 + 11500 x 0)/(sum of all frequencies)
You can perform this calculation to find the mean for the fingerling catfish distribution. Repeat this process for the Distribution of Large Catfish histogram to find its mean as well.
Once you have the means for both distributions, you can compare them to determine the best comparison of the means and the ranges for the two distributions.